Answer:
N=35%. so 35/100*7 =2.45%
H=5.0 so 5/100*1=0.05%
o=60% so 60/100*8=4.8%
Answer:
The molecules of the gas has more kinetic energy and has a lot of space between them. The molecules in the liquid moves slower than gas and has a tighter space than gas. Another difference is liquid takes up the shape of its container while gas does not. Cause its air.
Answer:
10.3 g Al
Explanation:
To find the excess mass of Al, you need to (1) convert grams I₂ to moles I₂ (via molar mass), then (2) convert moles I₂ to moles AlI₃ (via mole-to-mole ratio from equation coefficients), then (3) convert moles AlI₃ to grams AlI₃ (via molar mass). Now that you have the actual amount of AlI₃ produced, you need to (4) convert grams AlI₃ to moles AlI₃ (via molar mass), then (5) convert moles AlI₃ to moles Al (via mole-to-mole ratio from equation coefficients), and then (6) convert moles Al to grams Al (via molar mass). Now that you know the amount of Al actually needed to produce the product, you need to (7) find the excess mass of Al.
Molar Mass (Al): 26.982 g/mol
Molar Mass (I₂): 2(126.90 g/mol)
Molar Mass (I₂): 253.8 g/mol
Molar Mass (AlI₃): 26.982 g/mol + 3(126.90 g/mol)
Molar Mass (AlI₃): 407.682 g/mol
2 Al(s) + 3 I₂(g) -------> 2 AlI₃(g)
113 g I₂ 1 mole 2 moles AlI₃ 407.682 g
------------- x ---------------- x ---------------------- x ------------------- = 121 g AlI₃
253.8 g 3 moles I₂ 1 mole
121 g AlI₃ 1 mole 2 moles Al 26.982 g
--------------- x ------------------ x --------------------- x ----------------- = 8.01 g Al
407.682 g 2 moles AlI₃ 1 mole
Starting Amount - Mass Needed = Excess
18.3 g Al - 8.01 g Al = 10.3 g Al
for 39g water solute dissolved at 20C = solubility ( g/ 100 g H2O ) × mass of water = ( 11g / 100g H2O ) × 39g H2O = 4.29 g
amount of solute dissolved at 30 C =
= 23 / 100 * 39 = 8.97 g
Amount of extra solute dissolved = 8.97 - 4.29 = 4.7 g